Question: What is the value of the following logarithm? $\log_{6} \left(\dfrac{1}{216}\right)$
Explanation: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $6^{y} = \dfrac{1}{216}$ In this case, $6^{-3} = \dfrac{1}{216}$, so $\log_{6} \left(\dfrac{1}{216}\right) = -3$.